A276451 Number of 2-orbits of the cyclic group C_4 for a bi-colored square n X n grid with n squares of one color.

0, 1, 2, 12, 30, 408, 1012, 17920, 45600, 1059380, 2730756, 78115884, 203235032, 6917206576, 18113945256, 714851008512, 1881039165696, 84449819514060, 223049005408900, 11225502116862880, 29736777118603962, 1658138369930988088, 4403069737450280832

Offset: 1

Author

Chiang, Tung-Ying, Jason Y.S. Chiu, Hong-Chang Wang, Jiangshan Sun, Sep 03 2016

Keywords

• nonn
• easy

Comments

For a definition and examples of this problem see the comment section of A276449.

The present sequence a(n) gives the number of 2-orbits of such 2-color boards with n squares of one color under C_4.

Examples

			n = 4: one of the two 2-orbits is (o white, + black)
+ o + o   o o o +
o o o o   + o o o
o o o o   o o o +
o + o +   + o o o,
and one can take the first one as a representative.
For n = 3 there are a(3) = 2 2-orbits, represented by
+ o o       o o o
o + o  and  + + +
o o +       o o o.
The orbit structure for n=3 is 1^0 2^2 4^20; see A276449(3) = 0, a(3) = 2, A276452(3) = 20.
For the 12 2-orbits for n=4, see the representatives given in the link.
		

Links

• Hong-Chang Wang, Table of n, a(n) for n = 1..100
• Hong-Chang Wang, Example for n = 4

Crossrefs

Cf. A014062, A276449, A276452, A276454.

Programs

Mathematica

Table[(Function[j, Binomial[2 j (j + Boole@ OddQ@ n), j]]@ Floor[n/2] - If[MemberQ[{2, 3}, #], 0, Function[i, Binomial[(2 i) (2 i + #), i]]@ Floor[n/4]] &@ Mod[n, 4])/2, {n, 23}] (* Michael De Vlieger, Sep 07 2016 *)

Python

import math
def nCr(n,r):
    f = math.factorial
    return f(n) / f(r) / f(n-r)
# main program
for j in range(101):
    i = j/2
    if j%2==0:
        b = nCr(2*i*i,i)
    else:
        b = nCr(2*i*(i+1),i)
    if j%4==0:
        c = nCr((j*j/4),(j/4))
    elif j%4==1:
        c = nCr(((j-1)/2)*((j-1)/2+1),((j-1)/4))
    else:
        c = 0
    print(str(j)+" "+str((b-c)/2))

Formula

a(n) = (binomial(2*i*i,i) - A276449(n))/2, for n = 2*i.

a(n) = (binomial(2*i*(i+1),i) - A276449(n))/2, for n = 2*i+1.

Extensions

Edited: Wolfdieter Lang, Oct 02 2016